Question

A baseball is thrown horizontally off a 41-m high cliff with a speed of 18 m/s. What will be the angle in degrees between the horizontal direction and the velocity vector of the baseball just before it hits the ground? Assume that the answer will be positive and also will be the smallest angle between the horizontal and the velocity vector.

Answer 1

Horizontal component of velocity = 18 m/s

Vertical component of velocity, Vf, is given by
Vf²=Vi²+2*a*Δx
Vi=initial vertical velocity=0 m/s
a=acceleration due to gravity=-9.8 m/s²
Δx=-41 m

Vf=sqrt(0+2(-9.8)(-41))=28.4 m/s (downwards)
angle with horizontal
=atan(-28.4/18)

magnitude of velocity
=sqrt(-28.4²+18²)
=33.6 m/s